What is the integral of (4x^2-4x+6)/(x^3-x^2-6x) ?

The integral of (4x^2-4x+6)/(x^3-x^2-6x) is -ln|x|+3ln|x+2|+2ln|x-3|+C

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integral of (4x^2-4x+6)/(x^3-x^2-6x)

\int \frac{4 x ^{2} - 4 x + 6}{x ^{3} - x ^{2} - 6 x} d x

- ln ∣ x ∣ + 3 ln ∣ x + 2 ∣ + 2 ln ∣ x - 3 ∣ + C

Solution steps

\int \frac{4 x ^{2} - 4 x + 6}{x ^{3} - x ^{2} - 6 x} d x

Take the partial fraction of

\frac{4 x ^{2} - 4 x + 6}{x ^{3} - x ^{2} - 6 x} : - \frac{1}{x} + \frac{3}{x + 2} + \frac{2}{x - 3}

= \int - \frac{1}{x} + \frac{3}{x + 2} + \frac{2}{x - 3} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{1}{x} d x + \int \frac{3}{x + 2} d x + \int \frac{2}{x - 3} d x

\int \frac{1}{x} d x = ln ∣ x ∣

\int \frac{3}{x + 2} d x = 3 ln ∣ x + 2 ∣

\int \frac{2}{x - 3} d x = 2 ln ∣ x - 3 ∣

= - ln ∣ x ∣ + 3 ln ∣ x + 2 ∣ + 2 ln ∣ x - 3 ∣

Add a constant to the solution

= - ln ∣ x ∣ + 3 ln ∣ x + 2 ∣ + 2 ln ∣ x - 3 ∣ + C

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What is the integral of (4x^2-4x+6)/(x^3-x^2-6x) ?
The integral of (4x^2-4x+6)/(x^3-x^2-6x) is -ln|x|+3ln|x+2|+2ln|x-3|+C